A number divides 206, 368, and 449 leaving the same remainder in each cases.The largest such number is?

Question

anonymous · Answer

the answer is 81

apoorvk · Answer

Sorry I made a mistake. :/

okay let the remainder be 'x'. Now that basically means that if the three nos. had been (206-x), (368-x) and (449-x), they would have been exactly divisible by the divisor

anonymous · Answer

ok then

anonymous · Answer

Small typo,

(368 - 206, 368 - 449)=81

anonymous · Answer

@FoolForMath  how u got this (368 - 206, 368 - 449)=81 can u explain

anonymous · Answer

can u explain @FoolForMath

anonymous · Answer

I surely can, but the proof is somewhat big, sorry.

anonymous · Answer

can any one solve this????

anonymous · Answer

wait for asnaseer the big man!

anonymous · Answer

@FoolForMath  thanzz   for ur effort to help me..

asnaseer · Answer

I think this is the method what FFM is employing:

If 206 leaves some remainder after being divided by n, then we know:

206 - r = na

where r is the remainder and 'a' is the multiple of n.

therefore:

368 = 206 + 162 = na + 162

implies that 162 must also be exactly divisible by n.

similarly:

449 = 368 + 81 = na + 162 + 81

implies 81 must also be exactly divisible by n

asnaseer · Answer

therefore, the largest such multiple must be 81

asnaseer · Answer

hope it makes sense shameer1

anonymous · Answer

thznzzz

asnaseer · Answer

yw - and thx to FFM for introducing me to yet another new concept :)